Optoisolator issues on Current Transfer Ratio CTR

[matusel]

Hello boarders.

I'm analyzing a board wich uses a SHARP opto-isolator model PC357 and I have some issues with the datasheet. If you have a look at this link of datasheet:

http://sharp-world.com/products/device/lineup/data/pdf/datasheet/pc357n_e.pdf

On page 5 they show you a table in which PC357 optos are divided by rank on CTR (Ic/If) @ 5 milliamps f "If" .

Now, if you go on top-left of page 7, they show you a diagram of the CTR vs forward voltage; I can't explain how this diagram is related with the table on page 5 because there is no comment or relationship to the CTR-rank of the opto.

IMHO has not big sense add a similar diagram, seems just to give an idea of the curve trend of CTR vs If. BUT CAUSES CONFUSION.

Am I mistaken?
Do you have any suggestion on how can I obtain the theorical CTR @ 2.5mA, having a fixed rank (say for example RANK D) ?

 

[erikl]

Am I mistaken?

No, but I understand your confusion.

Fig.7 shows just a generic context between CTR and I_F, valid for all ranks. It tells you that - if the CTR is 270 at I_F=5mA, it will be 230 at I_F=2.5mA. So you can rank down the given D rank I_C limits (15 to 30mA) by this factor 230/270 .

BTW: V_CE doesn't have a strong effect on CTR (as long as V_CE >≈ 2V and IC ≦ 5mA, s. Fig.8), but the temperature effect (both on I_F and I_C, i.e. on CTR) isn't negligible (s. Fig.9).

 

[matusel]

No, but I understand your confusion.

Fig.7 shows just a generic context between CTR and I_F, valid for all ranks. It tells you that - if the CTR is 270 at I_F=5mA, it will be 230 at I_F=2.5mA. So you can rank down the given D rank I_C limits (15 to 30mA) by this factor 230/270 .

BTW: V_CE doesn't have a strong effect on CTR (as long as V_CE >≈ 2V and IC ≦ 5mA, s. Fig.8), but the temperature effect (both on I_F and I_C, i.e. on CTR) isn't negligible (s. Fig.9).

Thank you for your answer; I am aware of the temperature curve.

So, if I understand correctly, the CTR of the RANK D for example (minimum is CTR=15/5=3) should be 3 if forward current is 5 milli amps but decreases to 3*(230/270)= 2.55 if driven at 2.5 milli amps ?

 

[erikl]

So, if I understand correctly, the CTR of the RANK D for example (minimum is CTR=15/5=3) should be 3 if forward current is 5 milli amps but decreases to 3*(230/270)= 2.55 if driven at 2.5 milli amps ?

Right! 2.56 , if correctly rounded ;-)